Hello,(adsbygoogle = window.adsbygoogle || []).push({});

I posted a similar question long time ago, but after working on it I am still unable to arrive at a solution.

Let's have a group of linear transformations (rotations in thexy-plane):

[tex]R_\theta=\{ (\begin{array}{ccc} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}) \\ : \\ \theta \in [0,2\pi] \}[/tex]

The question is: How can I construct an orthogonal curvilinear coordinates system, in which the parameter [itex]\theta[/itex] works as one coordinate?

What I am supposed to get as a result are essentially the equations defining the cartesian-to-polar transformation.

----------------

My attempt:

Observe that given any vectorx, the orbit [tex]R_{\theta}(\mathbf{x})[/tex] is a parametric curve which is obviously a circle.

The (gradient) vectors

[tex]e_\theta=\frac{\partial R_{\theta}(\mathbf{x})}{\partial \theta}[/tex] are tangent to the curve, so if we consider their orthogonal complement [tex]e_\theta^*[/tex] (which is easy to find), we have already found a family oflocalorthogonal bases.

How can I continue from this point???

I am supposed to get: [itex]r = (x^2 + y^2)^{1/2}[/itex] and [itex]\theta = atan2(y/x) [/itex], but I don't know how to arrive at that.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Polar coordinates: derivation from rotation group

**Physics Forums | Science Articles, Homework Help, Discussion**